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## Geologic Time: Radiometric Time Scale

Half Life Chemistry Problems - Nuclear Radioactive Decay Calculations Practice Examples

Because the radioactive half-life of a given radioisotope is not affected by temperature, physical or chemical state, or any other influence of the environment outside the nucleus save direct particle interactions with the nucleus, then radioactive samples continue to decay at a predictable rate and can be used as a clock. This makes several types of radioactive dating feasible. For geologic dating, where the time span is on the order of the age of the earth and the methods use the clocks in the rocks , there are two main uncertainties in the dating process:. Starting with the simplest case where there are no daughter atoms present and no mass is lost from the sample, the age can be determined by measuring the relative amounts of the isotopes. This can be done by chemical means, but for precise determinations, mass spectrometry can be used.

This makes several types of radioactive dating feasible. For geologic dating, where the time span is on the order of the age of the earth and the methods use the clocks in the rocksthere are two main uncertainties in the dating process:. Starting with the simplest case where there are no daughter atoms present and no mass is lost from the sample, the age can be determined by measuring the relative amounts of the isotopes. This can be done by chemical means, but for precise determinations, mass spectrometry can be used.

From the radioactive decay equations, an expression for elapsed time can be developed. Using the common nuclear practice of calling the isotopes "parent" and "daughter", we use P and D to indicate the associated numbers of atoms. The requirement of keeping the same number of nuclei gives. Now suppose that there was an original amount of the daughter element present at the formation time of the sample being studied. This adds an additional unknown in the process, and requires an additional piece of data to permit a solution for elapsed time.

The requirement on the populations is now. Fortunately for radioactive dating processes, additional information is available in the form of other isotopes of the elements involved in the radioactive process. If there is another isotope of the daugther element D' which is presumed to be constant throughout the process, then the population requirement can be expressed in terms of the ratios. We can be reasonably confident that the isotope D' is contant if it is not radioactive not part of one of the natural radioactive series.

This makes several types of radioactive dating feasible. For geologic From the radioactive decay equations, an expression for elapsed time can be developed.

Using the radioactive decay equation as above, this becomes. Whatever we started with, we're going to have half left after 1. Divide both sides by N0. And then to solve for k, we can take the natural log of both sides. The natural log is just saying-- to what power do I have to raise e to get e to the negative k times 1. So the natural log of this-- the power they'd have to raise e to to get to e to the negative k times 1. Or I could write it as negative 1.

Radiocarbon dating is a method for determining the age of an object containing organic . The equation governing the decay of a radioactive isotope is: N = N 0 . During natural radioactive decay, not all atoms of an element are Using the equation below, we can determine how much of the original. Debunking the creationist radioactive dating argument. element still remaining in a mineral, it would be a simple matter to calculate its age by the formula.

That's the same thing as 1. We have our negative sign, and we have our k. And then, to solve for k, we can divide both sides by negative 1.

And so we get k. And I'll just flip the sides here. And what we can do is we can multiply the negative times the top. Or you could view it as multiplying the numerator and the denominator by a negative so that a negative shows up at the top. And so we could make this as over 1. It's just 1. Let me write it over here in a different color. The negative natural log-- well, I could just write it this way. If I have a natural log of b-- we know from our logarithm properties, this is the same thing as the natural log of b to the a power.

And so this is the same thing. Anything to the negative power is just its multiplicative inverse. So this is just the natural log of 2. So negative natural log of 1 half is just the natural log of 2 over here.

So we were able to figure out our k.

It's essentially the natural log of 2 over the half-life of the substance. So we could actually generalize this if we were talking about some other radioactive substance. And now let's think about a situation-- now that we've figured out a k-- let's think about a situation where we find in some sample-- so let's say the potassium that we find is 1 milligram. I'm just going to make up these numbers. And usually, these aren't measured directly, and you really care about the relative amounts. But let's say you were able to figure out the potassium is 1 milligram. And let's say that the argon-- actually, I'm going to say the potassium found, and let's say the argon found-- let's say it is 0. So how can we use this information-- in what we just figured out here, which is derived from the half-life-- to figure out how old this sample right over here?

How do we figure out how old this sample is right over there? Well, what we need to figure out-- we know that n, the amount we were left with, is this thing right over here. So we know that we're left with 1 milligram. And that's going to be equal to some initial amount-- when we use both of this information to figure that initial amount out-- times e to the negative kt. And we know what k is.

And we'll figure it out later. So k is this thing right over here. So we need to figure out what our initial amount is. We know what k is, and then we can solve for t. How old is this sample? We saw that in the last video. So if you want to think about the total number of potassiums that have decayed since this was kind of stuck in the lava.

And we learned that anything that was there before, any argon that was there before would have been able to get out of the liquid lava before it froze or before it hardened.

So maybe I could say k initial-- the potassium initial-- is going to be equal to the amount of potassium 40 we have today-- 1 milligram-- plus the amount of potassium we needed to get this amount of argon We have this amount of argon 0. The rest of it turned into calcium And this isn't the exact number, but it'll get the general idea. And so our initial-- which is really this thing right over here.